#include<stdio.h>
int main()
{
	int a[12] = { 1,2,3,4,5,6,7,8,9,10,11,12 }, * p[4], i;
	for (i = 0; i < 4; i++)
		p[i] = &a[i * 3];
	printf("%d\n",p[3][2]);
	return 0;
}
//将数组a表达为A[][3]
//即A[][3={{1,2,3},{4,5,6},{7,8,9},{10,11,12}}
//p为指针数组，存储着数组A的每行首元素地址，因此p[3][2]实际上为*(*(p+3)+2)=12